Question: Solve for $x$ and $y$ using elimination. $\begin{align*}8x+y &= 1 \\ 8x-y &= -5\end{align*}$
Explanation: We can eliminate $y$ when its corresponding coefficients are negative inverses. Add the top and bottom equations. $16x = -4$ Divide both sides by $16$ and reduce as necessary. $x = -\dfrac{1}{4}$ Substitute $-\dfrac{1}{4}$ for $x$ in the top equation. $8( -\dfrac{1}{4})+y = 1$ $-2+y = 1$ $y = 3$ $y = 3$ The solution is $\enspace x = -\dfrac{1}{4}, \enspace y = 3$.